Mathematical Modelling Lecture 2 -- Dimensional...

Dimensional AnalysisBuckingham’s Π-Theorem

Buckingham’s Π-Theorem ExampleSummary of Methodology

Mathematical ModellingLecture 2 – Dimensional Analysis

Phil [email protected]

Phil Hasnip Mathematical Modelling



Overview of Course

Model construction −→ dimensional analysisExperimental input −→ fittingFinding a ‘best’ answer −→ optimisationTools for constructing and manipulating models −→networks, differential equations, integrationTools for constructing and simulating models −→randomnessReal world difficulties −→ chaos and fractals

A First Course in Mathematical Modeling by Giordano, Weir &Fox, pub. Brooks/Cole. Today we’re in chapter 8.




Aim

To identify the relevant parameters and relationships forreal-world problems and hence guide experimental design




Functions of several variables

Suppose we have a function of x, y and z and we know that it islinear in x, y and z – i.e. if we fix y and z, and plot f against x weget a straight line; and the same if we fix x and z and vary y etc.How many different terms are there?

f (x) = mx + c

A straight line has 2 parameters (slope and intercept), and wehave 3 variables (x,y,z). Does this means we have 2× 3 = 6parameters in total?




Functions of several variables

f (x , y , z) = a1 + a2x + a3y + a4z + a5xy + a6xz + a7yz + a8xyz

We have 23 = 8.

This gets worse very quickly if it isn’t linear, but quadratic(33 = 27), cubic (43 = 64) or even higher order.

To get these parameters from experiment, we need at least oneexperimental measurement per parameter.




You are feeling very sleepy...

What affects the period τ of the pendulum?




You are feeling very sleepy...

Perhaps τ = f (l , m, g, θ). What is f?




Pendula

τ = f (l , m, g, θ)

Could use experiments to deter-mine f . How many measure-ments do we need?

Quadratic⇒ 3 parameters4 independent variables⇒ 34 = 81 totalparameters⇒ 81 expt measurements!




Dimensional analysis

P parameters per variable, V variables⇒ PV total parameters.

But what if we can reduce the number of variables that need tobe studied? Then we have a big saving!

Dimensional analysis does this by considering dimensionlessproducts.

NB Dimensions are not the same as units!





Dimensions M, L, T (and K, C, etc)Product (includes quotients) e.g. [area]=L2,

[energy]=ML2T−2, etc.Dimensionless product = combination s.t. dimensions are

M0L0T0




Dimensional compatibility

When adding terms in an equation they must all have the samedimension.

s = ut +12

at2

u is a velocity, LT−1, t is a time T⇒ ut is LT−1.T = L, i.e. a lengtha is acceleration, LT−2

⇒ at2 is also a length

You cannot add apples and oranges! Terms must bedimensionally compatible.




Dimensional homogeneity

The equation should be true regardless of units. This isachieved if the left- and right-hand sides have the samedimensions.

s = 12gt2

g is an acceleration LT−2, so 12gt2 is L

→ dimensionally homogeneouss = 4.6t2

Dimensionally inhomogeneous→ different answer ifmeasure time in seconds, minutes, hours...




Dimensionless products

Products of variables which are dimensionless are alwaysdimensionally homogeneous.

If a real world problem can be modelled by a dimensionallyhomogeneous equation (and no logarithms) then we can findthe form of that equation using dimensional analysis.




Pendulum Analysis

Variable Dimensionm Mg LT−2

τ Tl Lθ M0L0T 0

What are the dimensions of a general product mαgβτγ lδθε?

MαLβ+δT γ−2β




Pendulum analysis

MαLβ+δT γ−2β is dimensionless iff

M : α = 0L : β + δ = 0T : γ − 2β = 0

which gives an infinite set of solutions – not enough equations!

α = 0 =⇒ m cannot appear in the modelθ has no units =⇒ value is arbitrary




Dimensionless products

There are three basic rules when forming dimensionlessproducts:

1 Choose the dependent variable to appear once2 Choose any variable that always appears in each

dimensional equation3 Choose any variable that always has zero exponent (e.g. θ)




Pendulum – dimensionless product 1

Our dependent variable is τ , so choose γ = 1But γ − 2β =⇒ β = 1

2

δ = −β = −12

ε is arbitrary, so choose ε = 0

Thus our first dimensionless product is

Π1 = m0g12 τ l−

12 θ0 = τ

√gl




Pendulum – dimensionless product 2

Already have τ in first productFor second product choose γ = 0=⇒ β = 0=⇒ δ = 0ε arbitrary – choose ε = 1 (already used ε = 0)

Thus our second dimensionless product is

Π2 = m0g0τ0l0θ1 = θ




Pendulum analysis

Our dimensionless pendulum equation will relate the dps insome way

Π1 = f (Π2)

⇒ τ

√gl

= f (θ)

τ =

√lg

f (θ)

Quick check:

LHS has dimension TRHS has dimension

√L

LT−2 = TEquation is dimensionally homogeneous




Pendulum analysis

τ =

√lg

f (θ)

What have we learned?

τ does not depend on m ⇒ ‘only’ 53 = 125 experimentschanging units of time cannot change the actual period τ –there’s a corresponding change in ‘g’⇒ equation isdimensionally homogeneous.

We shall make this a bit more rigorous in a moment.

But for now...Phil Hasnip Mathematical Modelling



Pendulum analysis

τ =

√lg

.h (θ)

If we keep θ = θ0 = constant and vary l then

τ1

τ2=

√l1l2

i.e.

τ ∝√

l regardless of h.⇒ graph of τ against

√l should be linear

⇒ simple test requiring only 5 points!

If test fails, we go back and check our assumptions...Phil Hasnip Mathematical Modelling



Pendulum analysis

What about h (θ)? Fix l = l0 and vary θ:

τ1

τ2=

h (θ1)

h (θ2)

⇒ plot a graph of τ vs. θ (or better, τ√

gl vs. θ)

Can get h (θ)directly! Another insight gained ...

NB Whilst can do SHO analytically, cannot do the general casefor arbitrary θ as it is non-linear ...




A pendulum





Original problem had 1 dependent and 4 independentvariablesWe had 3 dimensional constraintsHence need 5− 3 = 2 dimensionless productsResult was an equation determined up to an unknownfunction of 1 dimensionless product




Buckingham’s Π-Theorem

A problem with n variables and m independent dimensionalconstraints can be written in dimensionally homogeneousform using (n −m) dimensionless products (dps) as

f (Π1,Π2, . . . Πn−m) = 0




Example 1

n = 4, m = 3

⇒ need 1 product including the dependent variable⇒ f (Π1) = 0 so can solve to get Π1 = constant⇒ can get dependent variable = (unknown constant) ×(other variables)




Example 2

n = 5, m = 3

⇒ need 2 products⇒ f (Π1,Π2) = 0can solve to get Π1 = h (Π2)

⇒ can get equation up to an unknown function of a singledimensionless product, as with the pendulum




Example 3

n = 6, m = 3

⇒ need 3 products⇒ f (Π1,Π2,Π3) = 0so can solve to get Π1 = h (Π2,Π3)

⇒ can get equation up to an unknown function of twodimensionless products, etc.

We always need to construct (n-m) independent dps, with thedependent variable only appearing once, e.g. in Π1.

NB Good to put the most sensitive variables into the dp with theindependent variable (e.g. Π1) to minimise the amount ofunknown behaviour and simplify experiments.




Buckingham’s Π-Theorem Example

Predict the period of 2 masses (m1 & m2) orbiting eachother at a distance R apart, in vacuum

Identify variables:

Variable Dimensionsτ T

m1 Mm2 MR LG L3M−1T−2

So we have 5 variables and 3 dimensions⇒ need 2 dps





General form of dp:

Π = ταmβ1 mγ

2RδGε

= T αMβMγLδL3εM−εT−2ε

= Mβ+γ−εL3ε+δT α−2ε

i.e. coefficients:T : α− 2ε = 0M : β + γ − ε = 0L : δ + 3ε = 0





T : α− 2ε = 0M : β + γ − ε = 0L : δ + 3ε = 0

Π1: include τ once⇒ α = 1⇒ ε = 12 ⇒ δ = −3

2 ⇒ β + γ = 12

so free choice, e.g. β = 1/2 and γ = 0

⇒ Π1 = τm1/21 R−3/2G1/2 = τ

√m1GR3





T : α− 2ε = 0M : β + γ − ε = 0L : δ + 3ε = 0

Π2 : set α = 0⇒ ε = 0⇒ δ = 0⇒ β + γ = 0 so free choiceexcept must not choose same as before, e.g. β = 1 and γ = −1

⇒ Π2 =m1

m2





Hence f (Π1,Π2) = 0

⇒ τ =

√R3

m1G.h

(m1

m2

)

Exact analytic answer: τ = 2π√

R3

G(m1+m2)

⇒ h(

m1

m2

)=

1√1 + m1/m2




Summary of Methodology

1 Decide your n variables, hence m dimensional constraints2 Form complete set of (n −m) dimensionless products

(dps):

dependent variable only appears once (e.g. in Π1)put most sensitive variables into same dpcheck each dp found has no dimensions!

3 Apply Buckingham’s Π-Theorem and hence solve fordependent variable

4 Test assumptions made (e.g. τ ∝√

l for pendulum)5 Conduct further experiments necessary to find any

unknown functions, or further computations based uponthe dps found.


Mathematical Modelling Lecture 2 -- Dimensional...

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